# Acidity order of o-nitrophenol vs. o-chlorophenol [closed]

It is given that the above statement is true. Why does o-nitrophenol, in spite of intramolecular hydrogen bonding, have a lower $$\mathrm{p}K_\mathrm{a}$$ than o-chlorophenol?

• The powerful inductive effect of the nitro group if far stronger than any H-bonding effect Nov 26, 2019 at 16:52
• @AniruddhaDeb Btw the above statement is true. o-chlorophenol has greater pKa (i.e. less acidic) than o-nitrophenol. Nov 26, 2019 at 16:55
• @Waylander Could you give a more rigorous answer? Are there any more examples of compounds where the inductive effect dominates over H-bonding? Or is o-nitrophenol an exception? Nov 26, 2019 at 16:55
• @SoumikDas edited accordingly. Nov 26, 2019 at 16:57

The thing you have to realise is that H-bonding can lower the ease with which the acidic $$\ce{H}$$ is removed, but it does not render it completely 'locked' within the molecule. What you have to understand is, the hydrogen bond shown here is not that strong as compared to the other kinds of H-bonds which have been observed in chemistry

Let us consider some values from Wikipedia:

Hydrogen bonds can vary in strength from weak ($$\pu{1–2 kJ mol−1}$$) to strong ($$\pu{161.5 kJ mol−1}$$ in the ion $$\ce{HF2−}$$. Typical enthalpies in vapor include:

$$\ce{F−H···:F}$$ ($$\pu{161.5 kJ/mol}$$ or $$\pu{38.6 kcal/mol}$$), illustrated uniquely by $$\ce{HF2−}$$, bifluoride

$$\ce{O−H···:N}$$ ($$\pu{29 kJ/mol}$$ or $$\pu{6.9 kcal/mol}$$), illustrated water-ammonia

$$\ce{O−H···:O}$$ ($$\pu{21 kJ/mol}$$ or $$\pu{5.0 kcal/mol}$$), illustrated water-water, alcohol-alcohol

$$\ce{N−H···:N}$$ ($$\pu{13 kJ/mol}$$ or $$\pu{3.1 kcal/mol}$$), illustrated by ammonia-ammonia

$$\ce{N−H···:O}$$ ($$\pu{8 kJ/mol}$$ or $$\pu{1.9 kcal/mol}$$), illustrated water-amide

$$\ce{HO−H···:OH+3}$$ ($$\pu{18 kJ/mol}$$ or $$\pu{4.3 kcal/mol}$$)

And from here, value of bond dissociation energy of phenolic $$\ce{O-H}$$ bond is approximately $$\pu{83.3 kcal/mol}$$.

Now, look at the value of the $$\ce{O−H··:O:}$$ bond and think, that while an extra 5 kcal/mol is needed to release the acidic proton, it's not that different as compared to a relatively quite large amount of $$\pu{83.3 kcal/mol}$$, which you have to anyhow supply as a base amount to break the phenolic $$\ce{O-H}$$ bond. To cause appreciably high hindrance in releasing the acidic proton, there are far higher values of H-bonds available, up to $$\pu{38.6 kcal/mol}$$, which is actually becoming comparable to the half of $$\pu{83.3 kcal/mol}$$ (i.e. $$\pu{41.65 kcal/mol}$$).

Hence, the proton can still be released fairly easily in o-nitrophenol, and if you compare the stability of conjugate base with radically less stabilising substituents like that in o-chlorophenol (as Waylander said in the comments), you can see a vast difference in which the reaction will be pushed towards conjugate base in the former case, so after a little bit of initial difficulty (relatively) in proton releasing, Le Chatlier principle will start pushing o-nitrophenol to it's conjugate base by itself.

However, it would become difficult to compare in neck-to-neck cases with another substituent of the caliber of -$$\ce{NO2}$$ (for example, -$$\ce{F}$$)

• Remember Bond Dissociation Energies are a measure of homolysis, not heterolysis, as your link mentions. Nov 26, 2019 at 21:27
• @user55119 So,the values I have are not for the hydrogen bonds and phenolic O-H bond? Dec 1, 2019 at 20:29
• In your discussion below the chart the value of 83.3 kcal/mol, which is taken from your link, is the BDE(O-H) reported for phenol. It is the value for homolysis as the the authors explain. I have no comment on your arguments about acidity. Dec 1, 2019 at 20:48
• @user55119 Oh, okay, thanks for the info! I am also adding a link in relation to this which can help people looking at this answer Dec 1, 2019 at 21:08

$$\mathrm{p}K_\mathrm{a}$$ is inversely proportional to acidity of a compound similar to $$\mathrm{p}K_\mathrm{b}$$ which is inversely proportional to basicity as it is $$-\log K_\mathrm{a}$$ and $$-\log K_\mathrm{b}.$$

In this case nitro group is a electron-withdrawing and deactivates the phenyl group the most, whereas $$\ce{Cl}$$ is a moderate deactivating group but ortho, para directing group. Hence more of the tendency to polarise the $$\ce{OH}$$ bond and increasing acidity is with the nitro substituted phenol.

Phenols have higher acidities than aliphatic alcohols due to the aromatic ring's tight coupling with the oxygen and a relatively loose bond between the oxygen and hydrogen of the hydroxyl group directly attached to the aromatic carbocyclic nucleus. For this reason, electron withdrawing groups (EWG) and/or electron donating groups (EDG) attached to aromatic nucleus play a significant role in acidity of phenols.

Phenols are more acidic when the ring is substituted with a EWG regardless of the position. For example, unsubstituted phenol has a $$\mathrm {p}K_\mathrm{a}$$ of approximately $$10$$. However, the acidity of each of o-, m-, and p-chlorophenol shows higher acidity than that of phenol ($$\mathrm {p}K_\mathrm{a}$$s are $$8.56, 8.97$$, and $$9.14$$, respectively) due to the negative inductive effect ($$\mathrm{−I}$$) of chlorine atom (the closer the $$\ce{Cl}$$ atom to oxygen of phenol, the bigger the inductive effect). Some GWGs has higher effect than that induced by inductive effect alone. The best example of this kind is nitrophenols. For instance, o-nitrophenol, m-nitrophenol, and p-nitrophenol have $$\mathrm {p}K_\mathrm{a}$$ values of $$7.2, 8.36,$$ and $$7.15$$, respectively. This higher acidity is due to additional negative mesomeric (resonance) effect ($$\mathrm{-M}$$) of nitro group compared to that of chloro group (which is actually shows some $$(\mathrm{+M})$$-effect). The extra resonance stabilization of conjugate base (phenoxide ion) is possible only when nitro group is at 2- or 4-position of the phenol. $$\mathrm{M}$$-effect is not possible when nitro group is at 3-position of the phenol (it is only $$(\mathrm{-I})$$-effect that make 3-nitrophenol more acidic than phenol).

Keep in mind that $$\pm$$-mesomeric effect is essentially same for ortho- and para-cases while inductive effect magnitude is distance dependent. For example, it would be dominant when nitro/chloro group is present in ortho-position rather than para-position. Based on those facts, we can conclude followings:

• Magnitude of $$(\mathrm{-I})$$-effect of nitro group ($$(\mathrm{-I})_\ce{NO2}$$) is larger than that of chloro group ($$(\mathrm{-I})_\ce{Cl}$$), based on lower $$\mathrm {p}K_\mathrm{a}$$ of m-nitrophenol ($$8.36$$) compared to $$\mathrm {p}K_\mathrm{a}$$ of m-chlorophenol ($$8.97$$): The acidity of meta-substituted phenols are solely based on $$(\mathrm{I})$$-effect.
• Magnitude of $$(\mathrm{+M})$$-effect of chloro group ($$(\mathrm{+M})_\ce{Cl}$$) is smaller than magnitude of $$(\mathrm{-I})$$-effect of chloro group ($$(\mathrm{-I})_\ce{Cl}$$) when it is at para-position: This conclusion is based on the comparison of $$\mathrm {p}K_\mathrm{a}$$ of p-chlorophenol ($$9.14$$) and that of phenol ($$10$$).
• Hence, magnitude of $$(\mathrm{-M})$$-effect of nitro group ($$(\mathrm{-M})_\ce{NO2}$$) is much larger than magnitude of $$(\mathrm{+M})$$-effect of chloro group ($$(\mathrm{+M})_\ce{Cl}$$): This conclusion is based on above two conclusions and the comparison of $$\mathrm {p}K_\mathrm{a}$$ of o-nitrophenol ($$7.2$$) and that of o-chlorophenol ($$8.56$$).

Thus, when both groups are at ortho-position, acidity of o-nitrophenol (with $$(\mathrm{-M-I})_\ce{NO2}$$) should be higher than that of o-chlorophenol (with $$(\mathrm{+M-I})_\ce{Cl}$$).